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Rotational Inertia
Purpose: To determine the moment of inertia of a rigid body.
Equipment: Moment of inertia apparatus, weights and hanger, stopwatch, two meter stick, micrometer caliper,
and 30 cm ruler.
Theory:
In this lab we will be studying the moment of inertia of two rotating disks about a fixed shaft. From the
Parallel Axis Theorem we know that the moment of inertia of a body rotating at a distance R from the center of
mass of the body is given by I  I CM  MR 2 , where M is the total mass of the object. In this experiment you
will place weights symmetrically on a rotating shaft. The total moment of inertia in this case will be the sum of
the moments of inertia of the weights and the shaft. Letting ICM be the moment of inertia of the rotating weight
assembly and Is that of the shaft, we expect the total moment of inertia to be I  MR 2  I CM  I s . Thus, we can
write I  MR 2  I 0 , where I 0  I CM  I s . A plot of I vs R2 will be a straight line with slope M and intercept I0.
We can determine the moment of inertia I of the system by dropping a weight attached to a string, which is
wrapped around a shaft. This will cause the system to begin rotating. We can time fall of the object and use this
 gt 2f

2
to compute the moment of inertia as I  mrs 
 1 , where m is the mass of the descending weight, h is the
 2h

initial height of the descending mass above the floor, t is the elapsed time for the fall, and rs is the radius of the
shaft. [The interested reader can read the derivation of this formula at the end of the lab.]
Instructions:
1. Place four 100 g weights on the horizontal shaft using the wing nuts. Adjust the wing nuts so that the center
of the two weights on each side is 15 cm from the center of the vertical shaft. Record this as R is the data
table.
2. Measure the diameter of the bare just below the point where the string is attached to the shaft using the
micrometer caliper.
3. Place 50 grams on the weight hanger and hang from the loop on the string. Wind the string on the shaft and
raise the hanger to it highest position. Make sure that the string winds evenly on the shaft and does not pile
up in one place.
4. Measure the diameter of the shaft plus the string where the string is wound on the shaft. Record this value
and calculate the effective radius as one half of the average between this diameter and the bare shaft
diameter.
5. Use the two-meter stick to measure the distance between the bottom of the weight hanger and the floor at the
hanger’s highest height.
6. Carefully hold one finger in front of one arm of the apparatus to keep it from rotating. Release the arm and
start the timer simultaneously. Take care to do this without imparting any initial velocity to the arm. Stop the
timer when the weight hits the floor. Record this time in the data table.
7. Repeat the experiment for the other positions in the table.
8. Calculate the moment of inertia for each set of data. In Excel, plot I vs R2. Fine the slope of the best fit line
to your data.
Data:
Mass of the descending weight (m): 0.10 kg.
Mass of rotating weights: 0.400 kg.
Total Mass of wing nuts: 0.056 kg.
Total rotating mass (M): 0.456 kg.
Bare shaft diameter
______________ Shaft plus string diameter
_______________
Effective radius of the shaft (rs) ______________ Height of the descending weight (h) _______________
R (m)
I (kg m2)
tf (sec)
R2 (m2)
0.15
0.13
0.11
0.09
0.07
0.05
Analysis:
Graph Data:
Slope ______________
Intercept ______________
Percent Error in slope ___________
What do the slope and the intercept represent?
Derivation of I:
2h
, where h is the height of the fall and t is the time elapsed.
t
Using the conservation of energy of the system, we have that the final kinetic energies of the falling mass and
1
1
that of the rotational motion is equal to the initial stored potential energy: mgh  mv 2f  I f2 . Since the
2
2
linear speed of the unraveling string is the same as that of the falling weight and v f  rs , we can replace
The average velocity for the falling weight is v f 
 gt 2f

 f and use the above value of v f to obtain I  mr 
 1 . Here, m is the mass of the descending weight, h is
 2h

the initial height of the descending mass above the floor, t is the elapsed time for the fall, and rs is the radius of
the shaft.
2
s